Question 1134893

you can use long division

Divide the leading coefficients of the numerator {{{x^8-1}}} and the divisor {{{x-2 }}}

{{{x^8/x=x^7}}}

multiply {{{x-2}}} by {{{x^7}}} : {{{x^8-2x^7}}}

subtract {{{x^8-2x^7 }}}from {{{x^8-1}}} to get {{{new}}} remainder
 
{{{x^8 - 1- x^8-2x^7=2x^7-1}}}


so far, you have {{{(x^8 - 1 )/(x - 2)= x^7+ (2x^7-1)/(x - 2)}}}


now repeat all with {{{2x^7-1}}}

{{{2x^7/x=2x^6}}}..........multiply {{{x-2}}}

{{{2x^6( x-2)=2x^7-4x^6}}}

subtract {{{2x^7-4x^6}}} from {{{2x^7-1}}}

{{{2x^7-1-(2x^7-4x^6)=2x^7-1-2x^7+4x^6=4x^6-1}}}

so far you have {{{x^7+2x^6+ (4x^6-1)/(x - 2)}}}

then {{{4x^6/x =4 x^5}}}

{{{4 x^5(x - 2)=4x^6-8x^5}}}

{{{4x^6-1-(4x^6-8x^5)=4x^6-1-4x^6+8x^5=8x^5-1}}}


and continue same process until you get 

{{{x^8 - 1 = (x^7 + 2 x^6 + 4 x^5 + 8 x^4 + 16 x^3 + 32 x^2 + 64 x + 128) * (x - 2) + 255}}}

=> reminder is {{{255}}}